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What percentage of the original kinetic energy is convertible?

  1. Oct 15, 2016 #1
    1. The problem statement, all variables and given/known data
    Ball 1 has an inertia of 0.500 kg and ball 2 has an inertia of 0.600 kg . Ball 1 is moving away from you at 5.0 m/s, and you decide to throw ball 2 at it to make it go faster. The balls collide head-on, and the coefficient of restitution for the collision is 0.95.
    Part A)How fast must ball 2 be traveling in order to double the velocity of ball 1?
    ANS: 9.7 m/s
    Part B) What is the initial relative velocity of the two balls?
    ANS: 4.7 m/s
    Part C)What is their reduced inertia?
    ANS: 0.27kg
    Part D )What percentage of the original kinetic energy is convertible?

    2. Relevant equations
    Kconv/Ki(x100%)

    3. The attempt at a solution
    Kconv/Ki (x100%)
    1.215/34.477 (x100%)= 3.5%
    ( every answer except Part D is correct, i'm looking for help with part D only)
     
  2. jcsd
  3. Oct 15, 2016 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Please show how you got the result for (d), otherwise it is impossible to tell what went wrong. I get a different value for Kconv, and that is more than a single step.
     
  4. Oct 15, 2016 #3
    Ki= (0.5)(0.500)(5)^2+(0.5)(0.600)(9.7)^2
    Ki=34.477

    Kconv=(0.5)(0.27)(9.7-5.0)^2
    =2.98251

    2.98251/34.477 (x100%)= 8.6%
     
  5. Oct 15, 2016 #4
    thats the correct answer, i found my error hours earlier and don't need help anymore
     
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